Posted by **Christina** on Monday, March 18, 2013 at 3:27pm.

The mean number of bumped airline passengers per day is 1.35 and the standard deviation is 0.25. For a random selection of a group of 40 days, what is the probability that the mean of bumped passengers for the 40 days will be between 1.25 and 1.50?

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**MathGuru**, Monday, March 18, 2013 at 6:02pm
Use this z-score formula for this problem:

z = (x - mean)/(sd/√n)

x = 1.25, 1.50

mean = 1.35

sd = 0.25

n = 40

Calculate two z-scores, then use a z-table to determine probability between the two scores.

I hope this will help get you started.

- stats. -
**Christina**, Monday, March 18, 2013 at 9:57pm
z=(1.25-1.35)/(0.25sqrt40)= -0.0791

z=(1.50-1.35)/(0325sqrt40)= 0.0791

but not sure how to read the z scores I don't know if it would be.5000-.5000 or what can someone please help me on the z scores for -0.0791 and 0.0791

- stats. -
**PsyDAG**, Tuesday, March 19, 2013 at 3:01pm
SD/√n = .25/6.3246 = .039528192 = .04

Z = (1.25-1.35)/.04 = -.10/.04 = -2.5

Z = (1.50-1.35)/.04 = .15/.04 = ?

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of both Z scores between those Z scores and mean. Add the two together.

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